A finite group $G$ is isomorphic to $\operatorname{Gal}(f,K)$.

Question

Let $G$ be a finite group with $1 \neq H < G $ a minimal subgroup which is not normal. Prove that there exists a field $K$ and a polynomial $f \in K[X]$ so that $G \cong \operatorname{Gal}(f,K)$ with $degf < |G|$.

From the comments:

I know that every finite G is isomorphic to some Galois extension. Let's say that every finite Galois extension is the splitting field of some polynomial (though I'm not sure about that). That's as far as I got.

Answer 1

Extended hints:

• Realize $G$ as a group of permutations on a finite set of variables $S=\{x_1,x_2,\ldots,x_n\}$. For example you can use the Cayley construction, when $n=|G|$. Often you get away with smaller $n$, but let's not worry about that.
• Let $L=\Bbb{Q}(S)$ be the purely transcendental extension of the rationals gotten be treating the elements of $S$ as algebraically independent variables. By the previous bullet identify $G$ as a (sub)group of automorphisms of $L$ simply by permuting the indeterminates $x_i$. Let $K$ be the fixed field of $G$, so by basic Galois theory the extension $L/K$ is Galois with Galois group $\cong G$.
• Let $F$ be the fixed field of $H$. Thus $K\subset F\subset L$, and $[L:F]=|H|$. Show that $F/K$ is a finite separable algebraic extension and thus simple. Conclude that $F=K(\alpha)$ for some $\alpha\in F$.
• Let $f(x)\in K[x]$ be the minimal polynomial of $\alpha$. Show that $L$ is the splitting field of $f$. Here is the beef. Clearly $L/K$ is Galois, but you need to show that no proper subfield $M$ such that $F\subseteq M\subset L$ is Galois. You need to make full use of the assumptions made about $H$ in this step.
• Convince yourself, your mates and your teacher that this does it.